Construction Review

University of Florida

Topic 1
Project Scheduling

University of Florida

1
Scheduling
Grouping of Activities
 Common constraints (ex. Load and non-load bearing walls)
• Labor
• Precedent activity
• Equipment
 Continuous tasks (ex. Floor slab and floor covering)
 Different attributes (excavation and placement and compaction of fill
materials and utility lines)
 Long Lead Items – Chillers

Types of Schedules
 Activity on Arrow
 Activity on Node
 Linear schedules
• Activities that are linear in nature
 Bar charts
• Software developed bar charts
• Simple bar chart
o Bar Charts do not show the relationship between activities.
o Which activities must be completed before an activity can start?
o Which activities are critical so that any delay to them will delay the
project?

2
Types of Project Scheduling Methods
Activity On Arrow (AOA) Diagrams
 Activity-on-Branch
 May have dummy tasks
 Finish to start precedence only

Activity on Node (AON) Diagrams

 Precedence Diagrams
 Can have any kind of precedence

Activity on Arrow
Diagrams

3
AOA Precedence Relationships
B
A
A must finish before either B or C can start

C
A
A and B must finish before C can start
C
B

A C
A and B must finish before either C or Dcan
start
B D

AOA Precedence Relationships

A B
Dummy

C D

A must finish before B can start;

Both A and C must finish before D can start

4
Dummy Activity Example
To be able to bolt a bracket to a panel, the
operations required are :

Which of the following is the correct AOA diagram to

disseminate the precedence above?

1 2 3
A A B A B
D
C C D C D

Critical Path Example

Activity Duration Precedence

A 3 -
B 6 A
C 10 A
D 11 A
E 8 B
F 5 C
G 6 D,F

5
Critical Path Example
The project sequence diagram is constructed:
Activity Duration Precedence
A 3 -
B 6 A
C 10 A
B D 11 A

A
C

Project
Start D

Critical Path Example

The remainder of the diagram constructed:
But; Now what? Activity Duration Precedence
E 8 B
F 5 C
G 6 D,F

E
E
B
A
C F
Project F
D G
Start

6
Critical Path Example

The project sequence graph is constructed:

E
B
A
C
F
Project G Project
Start D End

Critical Path Method (CPM) - Example

Activity times (duration) are added next :

B E
6 8
A C
3 10 F
Project D 5 G Project
Start 11 6 End

7
Critical Path Method (CPM) - Example

How do we find the critical path?

B E
6 8
A C
3 10 F
Project D 5 G Project
Start 11 6 End

Critical Path Method (CPM) - Example

The Critical Path is the path through the project on

which any delay will cause the completion of the
entire project to be delayed:

B E
6 8
A C
3 10 F
Project D 5 G Project
Start 11 6 End

8
Critical Path Method (CPM) - Example

The EARLIEST starting time of each activity is associated

with the events. It corresponds to the longest time of
any path from any previous event.
9

B E
6 8
0 A 3 C 13 24

3 10 F
Project D 5 G Project
Start 11 6 End
18

Critical Path Method (CPM) - Example

The LATEST starting time of each activity is also
associated with the events. It corresponds to the
longest time of any path from any subsequent event.
9
16
B E
6
8
0 A 3
C 13 24
0 3 10 13 24
3 F
Project D G Project
5
Start 11 6 End
18
18

9
Critical Path Method (CPM) - Example
The CRITICAL PATH is the path along which the earliest
time and latest time are the same for all events, and
the early start time plus activity time for any activity
equals the early start time of the next activity.
9
16 .
B E
6
8
0 A 3
C 13 24
0 3 10 13 24
3 F
Project D G Project
5
Start 11 6 End
18
18

Critical Path Method (CPM) - Example

This project cannot be completed in less than 24 weeks given the
expected duration of the activities. However, activities B, D and E
could be delayed or extended by 7, 4, and 7 weeks, respectively,
without affecting the minimum project completion time. This is termed
‘float’ or ‘slack’ time.
9
16
B E
6
8
0 A 3
C 13 24
0 3 10 13 24
3 F
Project D G Project
5
Start 11 6 End
18
18

10
 Critical Path

Activity Duration Earliest Latest Float

Start Start
A 3 0 0 0
B 6 3 10 7
C 10 3 3 0
D 11 3 7 4
E 8 9 16 7
F 5 13 13 0
G 6 18 18 0

Critical Path Method

Summary of Steps
 List all activities and expected durations.
 Construct CPM diagram for activities list.
 Determine EARLIEST start time for each event (working forward
from project start).
 Determine LATEST start time for each event (working backwards
from project end).
 Identify the CRITICAL PATH (and the ‘float’ time for any non-
critical activities).

11
 Arrow on Node
Diagrams

Logic Diagrams
Activity on Node, (AON)
Activities are represented as nodes and arrows represent the relationship
between activities. Complex relationships can be represented.

20 40
Build Wall Found. Build CMU Wall
Dur= 10 Dur= 15

10 60
Start Build ceiling beams
Dur= 0 Dur= 5

30 50
Build Col Found. Build Columns
Dur= 15 Dur= 15

12
AON Diagram Elements
Key Elements
• Early Start Date
• Late Start Date
• Early Finish Date
• Late Finish Date
• Duration

Types of Relationships
 Precedence
An activity follows another
 Concurrent
Activities can be performed at the same time
 Explicit
Activities that have a direct relationship. These are the only ones that
need to be represented.
 Simple
Activity B cannot start until activity A finishes (Finish to start, FS)

13
Complex Relationships

Start to Start, SS –
Activity B cannot start until
activity A starts

Finish to Finish, FF –
Activity B cannot finish
until activity A finishes

Lags
Refers to delays or leads added to the relationships.

L=2

Activity B cannot start until 2 days after activity A finishes.

L=-2

Activity B cannot start until 2 days before Activity A finishes.

14
Tasks Times
 Early Start, ES
The earliest time and activity can start given all constraints
 Early finish, EF
The earliest an activity can finish given all constraints
 Late start, LS
The latest time and activity can start without delaying the project
 Late finish, LF
The latest time and activity can finish without delaying the project.

Task Floats
 Total Float, TF
Critical activities have a total float of zero. Is the amount of time an
activity can be delayed without delaying the project.
TF=LF-Dur-ES
 Starting Float, SF
The start of a task can be critical without the whole activity being critical.
The difference between the late start and early start.
SF=LS-ES
 Finish Float, FF
The finish of a task can be critical without the whole activity being critical.
Difference between Late finish and early finish.
FF=LF-EF

15
Graphic Representation of PM

ES Act. # EF

Rel #
Activity Description and Duration Lag

LS TF LF

TF= LF –Dur-ES

Example 2 - AON
ES # EF

Des / Dur

20 LS TF LF
50
B 15 40 E 9 60
10
10

A 5 30 2
70
20 C 10 50
G 4
30 70
60

F 6 80
3
40

D 13

16
Example 2 - AON

Forward Pass

Begin the CPM Calculation by making a

forward pass. Start at the beginning and
calculate the early start and early finish for
each activity.

Example 2 - AON
Forward Pass ES # EF

EF=ES+Dur Des / Dur

5 20 20 LS TF LF
20 50 29
B 15 40 E 9 60
10
0 10 5

A 5 5 30 15 2 31 70 35
20 C 10 50
G 4
30 70
15 60 21

F 6 80
3
8 40 21

D 13

17
Example 2 - AON

Backward Pass

After completing the forward early date

calculation, begin with the last activity and work
backwards through the activities calculating the
late finish and late start dates.

Example 2 - AON
Backward Pass
ES # EF

LS=LF-Dur Des / Dur

5 20 20 LS TF LF
20 50 29
B 15 40 E 9 60
10 5 20
20 29
0 10 5

A 5 5 30 15 2 31 70 35
0 5 20 50
C 10 G 4
15 25 70
15 60 21 31 35
30

F 6 80
3
25 31
8 40 21

D 13
18 31

18
Example 2 - AON
Determining the Critical Path
ES # EF

I.e. - Activities with Zero Float Des / Dur

5 20 20 LS TF LF
20 50 29
B 15 40 E 9 60
10 5 0 20
20 0 29
0 10 5

A 5 5 30 15 2 31 70 35
0 0 5 20 50
C 10 G 4
30 15 25 70
15 60 21 31 0 35

F 6 80
3
25 31
8 40 21

D 13
18 31

Example 2 - AON
Calculating Total Float ES # EF

TF=LF-ES-Dur Des / Dur

5 20 20 LS TF LF
20 50 29
B 15 40 E 9 60
10 5 0 20
20 0 29
0 10 5

A 5 5 30 15 2 31 70 35
0 0 5 20 50
C 10 G 4
30 15 10 25 70
15 60 21 31 0 35

F 6 80
3
8 40 21
25 10 31 TFC=25-5-10
D 13 TFC=10
18 10 31

19
Example 2 - AON
Calculating Free Float ES # EF

FFAB= ESB-EFA Des / Dur

5 20 20 LS TF LF
20 50 29
B 15 40 E 9 60
10 5 0 20
20 0 29
0 10 5

A 5 5 30 15 2 31 70 35
0 0 5 20 50
C 10 G 4
30 15 25 70
15 60 21 31 0 35

F 6 80
3 FFDG=ESG-EFD
25 31
8 40 21
FFDG=31-21
D 13 FFDG=10
18 31

Topic 5
Earned Value Analysis

University of Florida

20
Earned Value Analysis
Definitions – Refer to the Manual
Earned Value Analysis
 BCWS – Budgeted cost of work scheduled (planned)
 ACWP – Actual cost of work performed (Actual)
 BCWP – Budgeted cost of work performed (Earned)
Variances
 Cost Variance (CV)
 Schedule Variance (SV)
 CV = PCWP – ACWP - - (Earned – Actual)
 SV = BCWP – BCWS - - (Earned – Planned)

Earned Value Analysis

Definitions – Refer to the Manual
Earned Value Analysis
• Cost Performance Index
– CPI = BCWP/ACWP

• Schedule Performance Index

– SPI = BCWP/BCWS
• Analysis
– CPI > 1.0  exceptional performance
– CPI < 1.0  poor performance
– Similar for SPI

21
Earned Value Analysis
PV – Planned Value or Budgeted Cost of Work Scheduled

BCWS – Budgeted cost of work scheduled (planned)

Earned Value Analysis

ACWP - Actual Cost of Work Performed

22
Earned Value Analysis
EV- Earned Value or Budgeted Cost of Work Performed

Earned Value Analysis

Actual Project Projections / Data “whole story”

23
Earned Value Analysis
Performance Metrics @ “June 03”
Schedule Performance Index
SPI: EV/PV = 49,000/55,000 = 0.891

Cost Performance Index

CPI: EV/AC = 49,000/56000 = 0.875

Earned Value Analysis - Example

Day “X”y

18

14

On Day X:
PLANNED VALUE (Budgeted cost of the work scheduled, BCWS)
BCWS = 18 + 10 + 16 + 6 = 50
EARNED VALUE (Budgeted cost of the work performed, BCWP)
BCWP = 18 + 8 + 14 + 0 = 40
ACTUAL COST (of the work performed , ACWP)
ACWP = 45 (from your project tracking - not evident in above chart)

24
Earned Value: Example

Cost (Person-Hours) Actual Cost: what you Today

have actually spent to
this point in time.

Planned Value: what your

plan called for sending on
the tasks planned to be
completed by this date.

Earned Value: value (cost)

of what you have
accomplished to date, per
the base plan.

Time (Date)

Earned Value: Example

Today
Cost (Person-Hours)

Over Budget

Behind
Schedule

Time (Date)

25
 Project Size Modifiers

University of Florida

Project Cost Modifiers:

Project Cost Modifiers are related to estimating:
For the purposes of the FE exam, the building type is located on page 176 of
the review manual and is assembled from RS means cost data based on:
building type and
relative size
First we need an initial estimate and then we work to modify it
Lets start off by using the table to get an idea of what how cost
estimates work:

26
Table from Reference book p.176

Table from Reference book p.176

The cost estimate for motel that is 18,000sf is most nearly?

18,000 $67.00/ $1,206,000? ?

27
Project Size Modifier Curve / Figure

Project Cost Modifier Figure:

Calculate size factor / multiplier:
Size factor = Proposed area/Typical area (or original area)
Back to our example = 18,000 /27,000 0.667

Multiplier = 1.06
Now apply to area

28
Project Cost Modifier Figure:
Modified cost estimate (based on size) :
∗
18,000 $67.00/ ∗ 1.06 = $1,278,360

The cost estimate for motel that is 18,000sf is most nearly?

a) $1,200,000
b) $1,206,000
c) $1,210,000
d) $1,280,000

Table from Reference book p.176

What is the cost estimate for an apartment that is 101,000sf?

29
 Extra Topic 1 – Earth Moving

University of Florida

Soil and Rock Properties

30
Types of Geotechnical Materials
 Gravel is rounded or semi-round
particles or rock that will pass a 3 in
and be retained on a 2.0 mm #10 sieve
 Sizes larger than 10 in are usually
called boulders
 Sand is disintegrated rock whose
particles vary in size from the lower
limit of gravel 2.0 mm down to 0.074
mm (#200 sieve).
 Sand can be classified as coarse or fine
sand, depending on the sizes of the
grains
 Silt is a material finer than sand, and
thus its particles are smaller than 0.074
mm but larger than 0.005 mm

Types of Geotechnical Materials

 Clay is a cohesive material whose particles are less than 0.005 mm
 Clays can be subject to considerable changes in volume with variations in
moisture content
 Organic matter is partly decomposed vegetation.
 Organic matter should be removed and replaced with a more suitable soil
 Borrow pit is a pit from which fill material is mined

31
Soils
 Bank cubic yards (bcy) In situ-
 Loose cubic yards (lcy)
 Compacted cubic yards (ccy)

bcy lcy ccy

Soils
 Bank cubic yards (bcy) In situ-
 Loose cubic yards (lcy)
 Compacted cubic yards (ccy)

32
Shrink and Swell Factors



– ∗
 %

.
 % ∗ 100
. –

Amount of Water Required

 The amount of water that must be added or removed is normally
computed in gallons per station (100 ft of length)

% – %
x

x
.

33
Application Rate
 The water application rate is normally calculated in gallons per square
yard

%
x

x ,

/
x
. /

Application methods
 Water Distributor

 Ponding

34
Application methods
 Sprinkling

Reducing the Moisture Content

 Drying actions may be as simple as aerating the soil

 They may, however, be as complicated as adding a soil stabilization agent

that actually changes the physical properties of the soil

 If a high water table is causing the excess moisture, some form of

subsurface drainage may be required before the soil’s moisture content
can be reduced

 The most common method of reducing moisture is to scarify the soil prior
to compaction

35
Reducing Moisture Content
 A Motor Grader Using Rippers to Scarify Material

Reducing Moisture Content

 Towed Disk Harrow used to scarify soil

36
Optimum Water Content
 The Optimum Water Content for compaction varies from about 12 to 25%
for fine-grained soils and from 7 to 12% for well-graded granular soils

 If the moisture content of a soil is below the optimum moisture range,

water must be added to the soil prior to compaction

 When processing granular materials, best results are usually obtained by

adding water in-place

 After water is added, it must be thoroughly and uniformly mixed with the
soil

Extra Topic 2
Earthmoving Operations

University of Florida

37
Machine Performance
 Cycle time and payload determine a machine’s production rate, and
machine travel speed directly affects cycle time.
 The three power questions:
1. Required Power
2. Available Power
3. Usable Power

Resistance
 Rolling Resistance (RR)
• The resistance of a level surface to constant-velocity motion

 Grade Resistance (GR)

• The force-opposing movement of a machine up a frictionless slope

 Total Resistance (TR)

• The power required is the power necessary to overcome the total
resistance to machine movement which is the sum of rolling and grade
resistance

38
Rolling Resistance
 Representative rolling resistances for various types of surfaces

Grade Resistance/Assistance
 The force-opposing movement of a machine up a frictionless slope is
known as grade resistance.

 It acts against the total weight of the machine, whether track type or
wheel type.

 When a machine moves up an adverse slope, the power required to keep

it moving increases approximately in proportion to the slope of the road.

 If a machine moves down a sloping road, the power required to keep it

moving is reduced in proportion to the slope of the road. This is known as
grade assistance.

39
Frictionless Slope-Force Relationships
 Components
sin
cos
• For angles less than 10°, sin (α) ≈ tan (α)
(the small –angle theorem) reduces to:

%
∗

• If we substitute W = 2,000 lb/ton, the formula reduces to:

20 / ∗ %

Total Resistance
 Total resistance can also be expressed as an effective grade.

/
%
/

40
Rim Pull Equation
 Rim Pull

375 ∗ ∗

 The efficiency of most tractors and trucks will range from 0.80 to 0.85
(use 0.85 if efficiency is not known).

Drawbar Pull
 The towing force a crawler tractor can exert on a load is referred to as
drawbar pull.

 Drawbar pull is typically expressed in pounds.

 To determine the drawbar pull available for towing a load it is necessary

to subtract from the total pulling force available at the engine the force
required to overcome the total resistance imposed by the haul conditions.

 If a crawler tractor tows a load up a slope, its drawbar pull will be reduced
by 20 lb for each ton of weight of the tractor for each 1% slope.

41
Performance Charts
 Equipment manufacturers publish
performance charts for individual
machine models.

 These charts enable the equipment

estimator/planner to analyze a
machine’s ability to perform under a
given set of project-imposed load
conditions.

 The performance chart is a graphical

representation of the power and
corresponding speed the engine and
transmission can deliver.

 The load condition is stated as either

rim pull or drawbar pull.

Performance Charts
 What is the maximum speed (and gear) for a grader
pulling/pushing a 25000 lb load?

42
Performance Charts
 What is the maximum speed (and gear) for a grader
pulling/pushing a 5000 lb load?

Performance Charts
 Performance Chart - Scraper

43
Example 1

Example #1: What is the maximum speed (and gear)

for a scraper which with Gross Vehicle Weight (GVW)
of 100,000 lb with 10% total resistance?

Performance Charts – Step 1

What is the maximum speed (and gear) for a scraper
which with Gross Vehicle Weight (GVW) of 100,000 lb
with 10% total resistance?
Start at the top
GVW = 100K

44
Performance Charts – Step 2
What is the maximum speed (and gear) for a scraper
which has Gross Vehicle Weight (GVW) of 100,000 lb
with 10% total resistance?
Draw a vertical line to
the total resistance
(10% in this case)

Performance Charts – Step 3

What is the maximum speed (and gear) for a scraper
which has Gross Vehicle Weight (GVW) of 100,000 lb
with 10% total resistance?

Draw a horizontal line to

the engine performance
(4th gear in this case)

45
Performance Charts – Step 4
What is the maximum speed (and gear) for a scraper
which has Gross Vehicle Weight (GVW) of 100,000 lb
with 10% total resistance?

Draw a vertical line to

the speed
(8MPH in this case)

Example 2

Example #2: What is the maximum speed (and gear)

for a scraper which with Gross Vehicle Weight (GVW)
of 22,000 lb with 10% total resistance?
(much lighter GVW than example 1)

46
Performance Charts – Example 2 - Step 1
What is the maximum speed (and gear) for a scraper
which has Gross Vehicle Weight (GVW) of 22,000 lb
with 10% total resistance?
Again, start at the top
GVW = 22K

Performance Charts – Example 2 - Step 2

What is the maximum speed (and gear) for a scraper
which has Gross Vehicle Weight (GVW) of 200,000 lb
with 10% total resistance?
Draw a vertical line to the
total resistance (10% in this
case)
Note: The intersection
between the total resistance
and the GVW is below the
performance curve.

47
Performance Charts – Example 2 - Step 3
What is the maximum speed (and gear) for a scraper
which has Gross Vehicle Weight (GVW) of 100,000 lb
with 10% total resistance?

Draw a horizontal line to

the engine performance
(8th gear in this case)
Note: The horizontal line is
in the opposite direction

Performance Charts – Example 2 - Step 4

What is the maximum speed (and gear) for a scraper
which has Gross Vehicle Weight (GVW) of 22,000 lb
with 10% total resistance?

Draw a vertical line to

the speed
(30 MPH in this case)

48
 Extra Topic 3
Earthwork Analysis and Mass Haul
Diagrams

University of Florida

Introduction

 Earthwork is the excavation of cut sections and placement of

soil in fill sections during construction.
 Earthwork Analysis
• Quantifies the amount of soil moved with consideration to
“haul” distance, borrow pits, and wasted soil
 Haul
• Transportation of material
 Borrow
• Excavating material from offsite to use as fill
 Waste
• Disposing of excavated material offsite

49
Excavation – Roadway Construction
 Includes:
Cut
Loading and Hauling
Placement in fill site
Compaction
Shaping for finished grade

Average End Area Method

• Average end area method is the most widely used
method to calculate the volume of soil between
stations in a roadway

50
Average End Area Method

Between Station 18+00 and 19+00, what is the excavated

volume in CY?

V = ((544 + 725)÷2) * (100÷27)

V = 2,350 yd3

Average End Area Method

What happens if the profile goes from cut to fill?

Station Area Distance Volume (ft3) Volume (yd3)

Number
22+00 350 --
23+00 100 100 22500 833.3
23+44 0 44 2200 81.5
24+00 -250 56 -7000 -259.3

The method requires that we account for the distance through

the portion of the profile with zero area

51
Haul Distance
Distance material is transported
From cut section to fill section
From borrow pit to fill section
From cut section to waste pit
Free Haul Distance (FHD):
Maximum distance material is transported by the
contractor without additional payment
Specified in construction contract
Cost incorporated into the excavation cost
Overhaul:
Any transported distance over the free haul distance

Overhaul - Example
Overhaul compensation is charged in units of sta-CY
or station-yards for the distance past FHD
Overhaul = CY * (haul distance – FHD)

Example: A contractor hauls 400 CY from a cut section

to a fill section 20 stations away. The construction
contract states that the FHD is 1200 feet. If the
excavation cost is $14/CY and the overhaul charge is
$2/ sta-yard, what is the overhaul cost?

52
Overhaul Problem
Overhaul cost = $2/sta-yards * 400 CY * ( 20 stations-12
stations)

Overhaul cost = $6400

What is the total cost?

Overhaul + Excavation cost = Total Cost

Excavation cost = $14/CY * 400 CY = $5600

Total Cost = $5600 + $6400 = $12,000

Limit of Economic Haul

 Limit of Economic Haul (LEH)
 The distance at which the overhaul distance equals borrow plus waste

 LEH < Haul distance

 Contractor hauls materials

 LEH > Haul distance

 Contractor uses borrow and waste pits

LEH = FHD + (borrow cost + waste cost – roadway excavation cost) /

overhaul cost

53
Limit of Economic Haul Problem
A contractor is determining the limit of economic haul
for a project. The free haul distance was determined to
be 800 feet. The borrow and waste costs are $22/CY
and $6/CY, respectively. The roadway excavation cost
is $12/CY and the overhaul cost is $2/sta-yard. What is
the LEH?

LEH = 8 stations + ($22/CY + $6/CY - $12/CY) / $2/sta-yard

LEH = 16 stations

Shrinkage Factors
 Compacted material takes up less volume than uncompact
material with the same mass
 A shrinkage factor (SF) is used to account for this volume
change
 A shrinkage factor only applies to fill sections

Volume of compacted fill = Volume of uncompact fill * [1 -

(SF/100) ]
Example:

Uncompact fill = 35 CY Volume of compact fill = 35 CY * [

SF = 15% 1- (15/100)]

= 29.75 CY needed to fill 35 CY

54
Rock Cuts
Rock cuts are an exception and do not need a
shrinkage factor applied to a fill section
Expands in fill section
Rock that is pulverized and blasted takes up more
volume

Earthwork Quantities
Construction quantities needed for various
construction items/operations
Clearing and Grubbing
Excavation
Overhaul
Borrow
Waste
Tables of quantities (itemized lists) are included in
plan drawings
Contractors use these quantities for bidding
Earthwork quantities incorporate shrinkage factors

55
Cut and Fill

Profile View

56
Mass Haul Diagrams

 Mass Haul Diagrams – an earthwork analysis method that

determines cut/fill volumes and haul distance to optimize
construction costs

Mass Haul Diagrams

 Mass Haul Diagrams illustrate:
 Amounts of earth excavation and embankment involved in balanced
sections
 Locations of balanced points
 Distribution of material
 Haul distances-
Haul distances and quantities combined enable section of
economical construction plant and scheduling of operations
 Direction of haul and limit of profitable haul distances
 Places where material can be borrowed or wasted to prevent
overhaul

57
Characteristics of Mass Haul Diagram

Characteristics of Mass Haul Diagrams

① Upward slope indicates excavation
② Downward slope indicated filling
③ Maximum earthwork point – end of excavation
④ Minimum earthwork point – end of an embankment
⑤ Difference between the transition points

 Whole volume of an embankment - vertical distance between

the maximum earthwork point and minimum earthwork point
 Whole volume of an excavation – vertical distance between
the minimum earthwork point and the maximum earthwork
point

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Characteristics of Mass Haul Diagrams
 Balance points – the point at which the prior excavated material can fill an
embankment
 Points a, c, and d
 Balancing lines - horizontal lines intersecting the mass curve show lengths
over which the cutting and filling are equalized
 Line xy quantifies the cut from X to B and the fill from B to Y
 The volume being moved is the vertical distance b to z
 Maximum haul distance – length of balancing line intercepted by the
mass curve
 If the curve lies above the balancing line, the material must be hauled
forward.
 If the curve lies below the balancing line, the material must be hauled
backward.

Excess vs. Borrow vs. Balanced

Borrow Excess Balanced

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Mass Diagram Creation - Revisited

Station Area Distance Volume Volume

Number (ft3) (yd3)
22+00 350 --
23+00 100 100 22500 833.3
23+44 0 44 2200 81.5
24+00 -250 56 -7000 -259.3

Characteristics of Mass Haul Diagram

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Overhaul by Methods of Moments
Steps
1. Create mass diagram
2. Plot cumulative yardage on y-axis
3. Plot stations on X-axis as the baseline
4. Negative (-) cumulative yardage is fill and plotted below
baseline. Positive (+) cumulative yardage is cut and plotted
above baseline.
5. Determine free haul distance on plot
6. Sum moments of overhaul about the stations where the
plotted free haul distance intercepts mass diagram
7. The total overhaul cost = (Sum of overhaul distances) X
(Overhaul yardage) X (Overhaul cost)

Overhaul - Method of Moments Example

The following table shows the stations and ordinates of a mass
haul diagram of a highway construction project. The free haul
distance is 500 feet. Overhaul cost is $5 per station-yard. Use
the method of moments to compute the total cost of overhaul.
Cumulative 0 800 950 1300 950 760 0
Yardage
Station 20+00 22+40 24+80 27+30 29+80 32+40 34+60

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 Overhaul - Method of Moments Example Cont’d

Free haul distance

Overhaul

Overhaul

Free Haul Distance = Center station ± 250’

= 27+30 ± 250’ = 24+80, 29+80

Acquisition of Moments about station 24+80 (left side)

= Sum Areas and divide by / Volume @ station

Overhaul - Method of Moments Example Cont’d

Free haul
A3 distance
A2
Overhaul

A1

Sum the Moments (left side):

[[(2240–2000)/2 + (2480 – 2240)]*(800) + [(2480-2240)/2]*(950-800)]/100

= 3,060 yd3-sta

Overhaul Distance = 3,060 yd3-sta ÷ 950 = 3.22 stations

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Overhaul - Method of Moments Example Cont’d

Free haul
Overhaul distance

Overhaul

Sum the Moments (left side):

[[(2240–2000)/2 + (2480 – 2240)]*(800) + [(2480-2240)/2]*(950-800)]/100

= 3,060 yd3-sta

Overhaul Distance = 3,060 yd3-sta ÷ 950 = 3.22 stations

Overhaul - Method of Moments Example Cont’d

Sum the Moments (right side):

[[(3460–3240)/2 + (3240 – 2980)]*(760) + [(3240-2980)/2]*(950-760)]/100

= 3,059 yd3-sta
Overhaul Distance = 3,059 yd3-sta ÷ 950 = 3.22 stations

Total Overhaul cost = (3.22 sta +3.22sta) * 950 yd3 * $5/sta-yd3

Total Overhaul cost = $30,590

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Mass Haul Example
The mass diagram represents the cut of 50,000-yd3
between station 40+00 and the upward balance point
at station 100+00 and a fill of 20,000-yd3 to the
upward transition point at station 130+00. Find the
average haul distance from station 40+00 through the
second transition point:

Mass Haul Example Cont’d - Profile and Mass Diagram

Direction of Haul

50,000 cy

25,000 cy 10,000 cy

20,000 cy

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